2 Answers
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If you just want to say "I've won the lottery", then all in a oner, provided you buy different numbers for the multiple tickets.

Suppose there are $Q$ possible outcomes in the lottery. The chance of winning with 1 ticket is $1/Q$. Assuming (reasonably) that the individual drawings are independent, the chance you fail to win $k$ times in a row is $(1-1/Q)^k$.

If you buy $k$ tickets at one time, the probability that the outcomes is not one of the $k$ tickets you bought is $(Q-k)/Q = 1 - k/Q$.

Observe that $(1-1/Q)^k > 1 - k/Q$ if $Q > 1$. So the chances of not having a winning ticket at all is greater if you play weekly. In other words, buying multiple tickets ever-so-slightly increase your chances of having a winning ticket. (The increase, however, is very, very small.)

But if you play for the pay-out: Assuming the pay-out is constant across the time-frame, let's call it $P$, buying $k$ tickets at once would give expected payout $kP/Q$: there's only one winning number, and the chance of you hitting it is $k/Q$.

Whereas if you play weekly, there is the chance that you win multiple times! The chances that you hit the jack pot $j$ times is $\binom{k}{j}(1-1/Q)^{k-j}(1/Q)^j$. But the payout for that would be $jP$. Adding them all up (from $j = 0$ to $j=k$, with a bit of algebra and the binomial theorem, you see that the total expected payout for playing weekly ends up also exactly as $kP/Q$, for playing $k$ times.